50 mL of a solution of Na2CO3 and NaHCO3 requires 17.5 mL of 0.1 M H2SO4 for neutralisation using phenolphthalein as indicator. Methyl orange is then added when further 25 mL of 0.2 M H2SO4 was required. Calculate the weight of Na2CO3 and NaHCO3 respectively in the solution.
The correct option is D 0.371 g, 0.546 g
Phenolphthalein will only neutralise 50% of Na2CO3.
12 Milliequivalents of Na2CO3=Milliequivalents of H2SO4=0.1×17.5×2=3.5
∴ Milliequivalents of Na2CO3=7.0
∴ WNa2CO3=7.0×1062×1000=0.371 g
The methyl orange indicator will neutralise the rest 50% of Na2CO3 and original NaHCO3.
∴ 12 Milliequivalents of Na2CO3+Milliequivalents of NaHCO3=Milliequivalents of H2SO4=0.2×25×2=10
∴ Milliequivalents of NaHCO3=10−3.5=6.5
∴ WNaHCO3=6.5×841×1000=0.546 g